A Square and a Cube — Class 8 Mathematics NCERT Solutions (Free)
Free step-by-step NCERT solutions for Class 8 Mathematics chapter "A Square and a Cube" — 8 important questions with detailed answers for CBSE board exam preparation.
TL;DR: Free step-by-step NCERT solutions for Class 8 Mathematics chapter "A Square and a Cube" — 8 important questions with detailed answers for CBSE board e…
Written & reviewed by the Syllab.in Academic Team (CBSE/NCERT subject experts) · Updated
NCERT Solutions for Class 8 Mathematics — A Square and a Cube. Step-by-step answers to all 8 textbook questions from this chapter, written for CBSE board preparation and free to use.
A Square and a Cube — All 8 Questions Solved
Q1. Find the square root of 256.
We need to find a number that when multiplied by itself gives 256.
Step 1: Let √256 = x, so x² = 256
Step 2: We can find this by prime factorization.
256 = 2 × 128 = 2 × 2 × 64 = 2 × 2 × 2 × 32
= 2 × 2 × 2 × 2 × 16 = 2 × 2 × 2 × 2 × 2 × 8
= 2 × 2 × 2 × 2 × 2 × 2 × 4 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2
= 2⁸
Step 3: √256 = √(2⁸) = 2⁴ = 16
Step 4: Verification: 16 × 16 = 256 ✓
Answer: √256 = 16
Q2. Which of the following is a perfect square? 64, 100, 128, 150
A perfect square is a number that can be expressed as the square of a whole number.
Step 1: Check 64
64 = 8 × 8 = 8²
✓ Perfect square
Step 2: Check 100
100 = 10 × 10 = 10²
✓ Perfect square
Step 3: Check 128
128 = 2 × 64 = 2 × 8²
This is not a perfect square because 128 = 2⁷, and the exponent is odd.
✗ Not a perfect square
Step 4: Check 150
150 = 2 × 75 = 2 × 3 × 25 = 2 × 3 × 5²
Since the exponents of 2 and 3 are odd, this is not a perfect square.
✗ Not a perfect square
Answer: 64 and 100 are perfect squares.
Q3. Find the cube root of 1000.
We need to find a number that when multiplied by itself three times gives 1000.
Step 1: Let ∛1000 = x, so x³ = 1000
Step 2: Prime factorization of 1000
1000 = 10 × 100 = 10 × 10 × 10 = 10³
or
1000 = 2 × 500 = 2 × 2 × 250 = 2 × 2 × 2 × 125
= 2³ × 125 = 2³ × 5³
Step 3: ∛1000 = ∛(10³) = 10
or
∛1000 = ∛(2³ × 5³) = ∛(2 × 5)³ = ∛(10³) = 10
Step 4: Verification: 10 × 10 × 10 = 1000 ✓
Answer: ∛1000 = 10
Q4. Is 216 a perfect cube? Justify your answer.
Step 1: Find the prime factorization of 216
216 = 2 × 108 = 2 × 2 × 54 = 2 × 2 × 2 × 27
= 2³ × 27 = 2³ × 3³
Step 2: Express as a product of prime factors
216 = 2³ × 3³ = (2 × 3)³ = 6³
Step 3: Check the exponents
For a number to be a perfect cube, all prime factors must have exponents divisible by 3.
Here, both 2 and 3 have exponent 3, which is divisible by 3.
Step 4: Verification: 6 × 6 × 6 = 36 × 6 = 216 ✓
Answer: Yes, 216 is a perfect cube because 216 = 6³
Q5. Find √(144/225).
We need to find the square root of a fraction.
Step 1: Use the property √(a/b) = √a/√b
√(144/225) = √144/√225
Step 2: Find √144
144 = 12 × 12 = 12²
√144 = 12
Step 3: Find √225
225 = 15 × 15 = 15²
√225 = 15
Step 4: Divide
√(144/225) = 12/15
Step 5: Simplify by finding GCD of 12 and 15
GCD(12, 15) = 3
12/15 = (12÷3)/(15÷3) = 4/5
Answer: √(144/225) = 4/5 or 0.8
Q6. Find the smallest number by which 75 must be multiplied to make it a perfect square.
Step 1: Find the prime factorization of 75
75 = 3 × 25 = 3 × 5²
Step 2: Analyze the exponents
For a perfect square, all prime factors must have even exponents.
- Prime factor 3 has exponent 1 (odd)
- Prime factor 5 has exponent 2 (even)
Step 3: Determine what to multiply
To make the exponent of 3 even, we need to multiply by 3 (to make it 3²).
The exponent of 5 is already even.
Step 4: The number to multiply is 3
75 × 3 = 225
Step 5: Verify that 225 is a perfect square
225 = 3² × 5² = (3 × 5)² = 15²
✓ It is a perfect square
Answer: The smallest number is 3
Q7. What is the smallest number by which 128 must be divided to make it a perfect cube?
Step 1: Find the prime factorization of 128
128 = 2 × 64 = 2 × 2 × 32 = 2 × 2 × 2 × 16
= 2 × 2 × 2 × 2 × 8 = 2 × 2 × 2 × 2 × 2 × 4
= 2 × 2 × 2 × 2 × 2 × 2 × 2 = 2⁷
Step 2: Analyze the exponents for perfect cube
For a perfect cube, all prime factors must have exponents divisible by 3.
- Prime factor 2 has exponent 7
Step 3: Find the remainder when 7 is divided by 3
7 = 3 × 2 + 1
Remainder is 1
Step 4: To make the exponent divisible by 3, we need to remove 2¹
Divide by 2¹ = 2
128 ÷ 2 = 64 = 2⁶ = (2²)³ = 4³
Step 5: Verify
64 = 4 × 4 × 4 = 4³ ✓
Answer: The smallest number is 2
Q8. If (x)² = 2401, find the value of x.
We need to find a number whose square is 2401.
Step 1: We need to find √2401
Step 2: Find the prime factorization of 2401
2401 = 7 × 343 = 7 × 7 × 49 = 7 × 7 × 7 × 7 = 7⁴
Step 3: Calculate the square root
√2401 = √(7⁴) = 7² = 49
Step 4: Verification
49 × 49 = 2401 ✓
Step 5: Consider both positive and negative roots
If x² = 2401, then x = ±49
Answer: x = 49 or x = -49
A Square and a Cube — Practice MCQs with Answers
Attempt these 20 multiple-choice questions after working through the solutions above. Each carries the correct option and the reasoning behind it, so a wrong answer tells you which idea to revisit.
MCQ 1. Which of the following numbers is a perfect square?
- A. A) 243
- B. B) 324
- C. C) 500
- D. D) 675
Answer: B. B) 324 — A perfect square is a number that can be expressed as the product of two equal integers. 324 is the square of 18 (18 × 18 = 324).
MCQ 2. What is the unit digit of the square of 78?
- A. A) 4
- B. B) 6
- C. C) 8
- D. D) 2
Answer: A. A) 4 — The unit digit of the square of a number is determined by the unit digit of the original number. The unit digit of 78 is 8, and 8² = 64, so the unit digit of 78² is 4.
MCQ 3. Which of these numbers cannot be the unit digit of a perfect square?
- A. A) 1
- B. B) 4
- C. C) 7
- D. D) 9
Answer: C. C) 7 — Perfect squares can only have unit digits 0, 1, 4, 5, 6, or 9. The digit 7 is not possible as a unit digit of any perfect square.
MCQ 4. How many non-perfect square numbers lie between 12² and 13²?
- A. A) 24
- B. B) 25
- C. C) 26
- D. D) 23
Answer: A. A) 24 — The number of non-perfect square numbers between n² and (n+1)² is 2n. Here, n=12, so 2 × 12 = 24.
MCQ 5. Find the square root of 729 using the prime factorization method.
- A. A) 27
- B. B) 29
- C. C) 31
- D. D) 33
Answer: A. A) 27 — Prime factorization of 729 is 3 × 3 × 3 × 3 × 3 × 3. Grouping them in pairs gives (3 × 3 × 3) × (3 × 3 × 3) = 27 × 27. Thus, the square root of 729 is 27.
MCQ 6. What is the smallest number by which 288 must be multiplied to make it a perfect square?
- A. A) 2
- B. B) 3
- C. C) 5
- D. D) 6
Answer: A. A) 2 — The prime factorization of 288 is 2⁵ × 3². To make it a perfect square, all prime factors must have even powers. The factor '2' has an odd power (5), so we need one more '2' to make its power even.
MCQ 7. The area of a square field is 961 m². What is the length of its side?
- A. A) 29 m
- B. B) 31 m
- C. C) 33 m
- D. D) 39 m
Answer: B. B) 31 m — The area of a square is side². To find the side length, we need to find the square root of the area. √961 = 31.
MCQ 8. Which of the following is a Pythagorean triplet?
- A. A) (3, 4, 6)
- B. B) (5, 12, 13)
- C. C) (7, 8, 9)
- D. D) (6, 9, 10)
Answer: B. B) (5, 12, 13) — A Pythagorean triplet (a, b, c) satisfies the condition a² + b² = c². For (5, 12, 13), 5² + 12² = 25 + 144 = 169, which is equal to 13².
MCQ 9. What is the sum of the first 9 odd natural numbers?
- A. A) 18
- B. B) 36
- C. C) 81
- D. D) 100
Answer: C. C) 81 — The sum of the first 'n' odd natural numbers is n². For n=9, the sum is 9² = 81.
MCQ 10. A general wants to arrange his 441 soldiers in a square formation. How many soldiers will be in each row?
- A. A) 19
- B. B) 21
- C. C) 23
- D. D) 25
Answer: B. B) 21 — If the soldiers are arranged in a square formation, the number of soldiers in each row is the square root of the total number of soldiers. √441 = 21.
MCQ 11. Without performing any calculation, which of the following numbers cannot be a perfect square?
- A. A) 1024
- B. B) 1296
- C. C) 1849
- D. D) 2357
Answer: D. D) 2357 — A perfect square can only have unit digits 0, 1, 4, 5, 6, or 9. The number 2357 ends with 7, which is not a possible unit digit for a perfect square.
MCQ 12. Find the smallest number by which 300 must be divided to get a perfect square.
- A. A) 2
- B. B) 3
- C. C) 4
- D. D) 6
Answer: B. B) 3 — Prime factorization of 300 is 2² × 3 × 5². To make it a perfect square, all prime factors must have even powers. The factor '3' has an odd power (1), so we must divide by 3.
MCQ 13. Find the square root of 1764 by the division method.
- A. A) 32
- B. B) 38
- C. C) 42
- D. D) 44
Answer: C. C) 42 — Using the division method for √1764: 40²=1600, 50²=2500. Since 1764 ends in 4, the root must end in 2 or 8. Checking 42 × 42 = 1764.
MCQ 14. If the side of a square is 1.5 cm, what is its area?
- A. A) 2.25 cm²
- B. B) 3.0 cm²
- C. C) 1.5 cm²
- D. D) 2.5 cm²
Answer: A. A) 2.25 cm² — The area of a square is side². So, (1.5 cm)² = 1.5 × 1.5 = 2.25 cm².
MCQ 15. Which of the following numbers is a perfect cube?
- A. A) 125
- B. B) 200
- C. C) 343
- D. D) Both A and C
Answer: D. D) Both A and C — A perfect cube is a number that can be expressed as the product of three equal integers. 125 = 5 × 5 × 5, and 343 = 7 × 7 × 7. Both are perfect cubes.
MCQ 16. What is the unit digit of the cube of 37?
- A. A) 1
- B. B) 3
- C. C) 7
- D. D) 9
Answer: B. B) 3 — The unit digit of the cube of a number is determined by the unit digit of the original number. The unit digit of 37 is 7, and 7³ = 343, so the unit digit of 37³ is 3.
MCQ 17. Find the cube root of 1331.
- A. A) 9
- B. B) 11
- C. C) 13
- D. D) 17
Answer: B. B) 11 — Using prime factorization, 1331 = 11 × 11 × 11. Therefore, the cube root of 1331 is 11.
MCQ 18. What is the smallest number by which 675 must be multiplied to make it a perfect cube?
- A. A) 5
- B. B) 3
- C. C) 9
- D. D) 15
Answer: A. A) 5 — Prime factorization of 675 is 3³ × 5². To make it a perfect cube, all prime factors must appear in groups of three. The factor '5' appears twice, so we need one more '5' to make 5³ a factor. So, multiply by 5.
MCQ 19. The volume of a cubical tank is 216 m³. Find the length of its side.
- A. A) 4 m
- B. B) 5 m
- C. C) 6 m
- D. D) 8 m
Answer: C. C) 6 m — The volume of a cube is side³. To find the side length, we need to find the cube root of the volume. ³√216 = 6.
MCQ 20. If the side of a cube is doubled, its volume increases by how many times?
- A. A) 2
- B. B) 4
- C. C) 6
- D. D) 8
Answer: D. D) 8 — Let the original side be 'a'. Original volume = a³. If the side is doubled, the new side is '2a'. New volume = (2a)³ = 8a³. So, the volume increases by 8 times.
How to Use These Solutions
Attempt each question yourself first and only then compare with the worked answer. Marks in the CBSE board exam are awarded for the METHOD as much as the final result, so reproduce the steps rather than memorising the last line. Where a solution states a law, a formula or a definition, learn that wording — examiners look for it.
Related practice for this chapter: chapter MCQs, previous-year questions, revision notes and sample papers.